How is the total number of electrons represented in a Lewis structure determined?

Mar 9, 2017

How else but by counting up (i) the number of valence electrons on the neutral atoms, and (ii) the charge on the atom or ion............?

Explanation:

Now we build Lewis structures by elaborating from neutral atoms, and of course, we have to account for the charge on the atom or radical ion.

Take for example $\text{nitrate ion}$, $N {O}_{3}^{-}$. Nitrogen, Group V, has 5 valence electrons; oxygen, Group VI, has 6 valence electrons. And we throw in another electron, so that we have $5 + 3 \times 6 + 1 = 24$ $\text{valence electrons}$, i.e. $\text{12 electron pairs}$ in the Lewis structure of $N {O}_{3}^{-}$ to distribute around 4 centres.

And thus we get $O = {N}^{+} {\left(- {O}^{-}\right)}_{2}$. From the left, around the doubly bound oxygen there are $2 + 2$ lone pair electrons on oxygen, (i) $4$ electrons, and in the $O = N$ bond, (ii) $4$ electrons, NO lone pairs on the cationic, quaternized nitrogen, and (iii) $2 \times 8 = 16$ electrons on the formally singly bound oxygens, each of which bears a NEGATIVE charge: thus 24 valence electrons as required for the Lewis structure.

So all we do is to take the number of valence electrons (which is given by the atom's Group number), and add these numbers together, and add or subtract depending on the negative or positive charge of the species.

Can you do the same for $P {O}_{4}^{- 3}$? There must be 32 electrons to distribute, i.e. 16 electron pairs.